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Math Help - ln and tanh

  1. #1
    Forum Admin topsquark's Avatar
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    ln and tanh

    Evaluate
    \displaystyle \int \frac{dy}{1 - y^2}

    There are essentially two ways to approach this. The first is
    \displaystyle \int \frac{dy}{1 - y^2} = tanh^{-1}(y) + C

    The other is to use partial fractions:
    \displaystyle \int \frac{dy}{1 - y^2} = \frac{1}{2} ln \left | \frac{1 + y}{1 - y} \right | + C

    Now, tanh^(-1) and ln have an overlapping domain and in this region everything is just fine and C = 0. The problem is that the ln solution has a much larger domain than the tanh one.

    How can two solutions of an integral have different domains?

    -Dan
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by topsquark View Post
    Evaluate
    \displaystyle \int \frac{dy}{1 - y^2}

    There are essentially two ways to approach this. The first is
    \displaystyle \int \frac{dy}{1 - y^2} = tanh^{-1}(y) + C

    The other is to use partial fractions:
    \displaystyle \int \frac{dy}{1 - y^2} = \frac{1}{2} ln \left | \frac{1 + y}{1 - y} \right | + C

    Now, tanh^(-1) and ln have an overlapping domain and in this region everything is just fine and C = 0. The problem is that the ln solution has a much larger domain than the tanh one.

    How can two solutions of an integral have different domains?

    -Dan
    Because is...

    \displaystyle \frac{1}{1-y^{2}}= \frac{1}{2}\ (\frac{1}{1+y}+ \frac{1}{1-y}) (1)

    ...You can in any case write...

    \displaystyle \int \frac{dy}{1-y^{2}} = \frac{1}{2}\ (\ln |1+y| - \ln |1-y|)+ c (2)

    ... and the expression (2) is valid both for |y|<1 and |y|>1. The alternative...

    \displaystyle \int \frac{dy}{1-y^{2}} = \tanh^{-1} y + c (3)

    ... is valid only for |y|<1...

    Kind regards

    \chi \sigma
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chisigma View Post
    Because is...

    \displaystyle \frac{1}{1-y^{2}}= \frac{1}{2}\ (\frac{1}{1+y}+ \frac{1}{1-y}) (1)

    ...You can in any case write...

    \displaystyle \int \frac{dy}{1-y^{2}} = \frac{1}{2}\ (\ln |1+y| - \ln |1-y|)+ c (2)

    ... and the expression (2) is valid both for |y|<1 and |y|>1. The alternative...

    \displaystyle \int \frac{dy}{1-y^{2}} = \tanh^{-1} y + c (3)

    ... is valid only for |y|<1...

    Kind regards

    \chi \sigma
    I realize that much. My question is that I had thought the integral of an expression was unique up to an arbitrary constant. Because the domains are different, the tanh^(-1) and ln are different functions. I know of no reason why the tanh^(-1) solution does not also represent the most general solution.

    -Dan
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  4. #4
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    I don't know why you would say "tanh^{-1}(x)[/tex] is the most general solution". As chisquared said, it is only valid for -1< x< 1. The ln solution is valid for |x|> 1. There are each the correct integral on that region.
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